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The heat of neutralization of $\ce{HCl(aq)}$ by $\ce{NaOH}$ is $\pu{-55.9 kJ/mol}$ $\ce{H2O}$ produced. If $\pu{50ml}$ of $\pu{1.6M}$ $\ce{NaOH}$ at $\pu{25.15^\circ C}$ is added to $\pu{25ml}$ of $\pu{1.79M}$ $\ce{HCl}$ at $\pu{26.34^\circ C}$ in a plastic foam cup calorimeter, what will be solution temperature be immediately after the neutralization reaction has occurred?

I'm always overwhelmed by the question if it provided me too much information. I don't know what equation and concept to use first. (But I know all related-equations in this unit.)

I need a detailed explanation to understand! I'm not sure which one is the initial temperature? $\pu{25.15^\circ C}$ of $\ce{NaOH}$ or $\pu{26.34^\circ C}~\ce{HCl}$?

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First, you need to recognize what the limiting reagent is. Because NaOH and HCl react in a 1:1 ratio, and there is less HCl ($\pu{25mL}$ of $\pu{1.79M}$ is fewer moles than $\pu{50 mL}$ of $\pu{1.6M}$), HCl is the limiting reagent.

Second, given the amount of limiting reagent, how much heat will be released (or absorbed). $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$

Third, you need to approximate that the solution has the heat capacity of water, which is $\pu{4.18 kJ/K L}$.

If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average. So here: $$(50 \times 25.15 + 25 \times 26.35)/75 = 25.55$$

So finally calculate the amount the temperature of $\pu{75 ml}$ of "water" will increase when $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$ of heat is added, and add this to $\pu{25.55 ^\circ C}$.

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    $\begingroup$ If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average. As long as there is no phase changes. $\endgroup$ – LDC3 Nov 18 '14 at 2:20

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